PDEs in Moving Time Dependent Domains

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PDEs in Moving Time Dependent Domains

Manuel Fernando Cortez, Aníbal Rodríguez-Bernal

To cite this version:

PDEs in moving time dependent domains

∗

Fernando Cort´ez

An´ıbal Rodr´ıguez-Bernal

Departamento de Matem´atica Aplicada Universidad Complutense de Madrid,

Madrid 28040, SPAIN and

† _{Instituto de Ciencias Matem´aticas}

CSIC-UAM-UC3M-UCM Dedicated to Professor M.G. Velarde

in occasion of his 70th birthdate.

Abstract

In this work we study partial differential equations defined in a domain that moves in time according to the flow of a given ordinary differential equation, starting out of a given initial domain. We first derive a formulation for a particular case of partial differential equations known as balance equations. For this kind of equations we find the equivalent partial differential equations in the initial domain and later we study some particular cases with and without diffusion. We also analyze general second order differential equations, not necessarily of balance type. The equations without diffusion are solved using the characteristics method. We also prove that the diffusion equations, endowed with Dirichlet boundary conditions and initial data, are well posed in the moving domain. For this we show that the principal part of the equivalent equation in the initial domain is uniformly elliptic. We then prove a version of the weak maximum principle for an equation in a moving domain. Finally we perform suitable energy estimates in the moving domain and give sufficient conditions for the solution to converge to zero as time goes to infinity

1

Introduction

In a standard setting for many partial differential equations of mathematical physics, one usually assumes that the physical process being described occurs in a fixed domain of the physical space. This includes many equations describing the motion of fluids for example,

despite the fact that particle fluids and hence fluid subdomains actually move with time. Of course there are some other problems,such as free boundary problems, in which the physical domain of the PDE changes with time. In all these problems the motion of particles or subdomains occurs according to an unknown velocity field with is actually one of the main unknowns of the problem.

In this paper we assume some intermediary situation in which each point of a given

initial domain Ω0 ⊂ Rn, moves in time according to some prescribed autonomous vector

field. Hence at later times the domain Ω0evolves into a diffeomorphic domain Ω(t) (which

is not excluded to coincide with Ω0 itself!). In particular, topological properties of the

domain are preserved along time. However the geometrical evolution of the domain can

be very complex; for example one can consider the evolution of the open set Ω0 in R3

with the vector field of the Lorenz equations in a chaotic regime.

Our goal is the to describe some sensible class of PDEs to be consider in such a family of moving domains. We choose then to describe balance equations in moving domains, which result from conservation principles and which have natural applications to conservation of mass, momentum, energy etc. For such equations one must then give some suitable definition of solution.

After giving a convenient meaning of solution for both balance and general parabolic equations, we prove that such equations can be solved using available results.

Then we investigate, on some particular, although significative examples of equations in moving domains, basic tools in the analysis of parabolic equations such as the (weak) maximum principle and energy estimates. In particular we obtain sufficient conditions on the equations and on the moving domains, that guarantee that the solutions converge to zero as time goes to infinity.

2

Moving domains

We assume that each point x of an original given domain (smooth open set) Ω0 ⊂ Rn,

starting at time t = 0 moves following a curve t 7−→ Y (t; x), in Rn_{. Moreover we assume}

this curve is a solution of the autonomous system of ODEs

 _˙

Y(t; x) = ~V(Y (t; x))

Y(0; x) = x (2.1)

for some given smooth vector velocity field ~V _{: R}n

−→ Rn_{. Even more and for simplicity}

we assume that all solutions of (2.1) are defined for all t ∈ R. Hence, for t ∈ R, we have a deformation map

φ(t) : Rn

−→ Rn_{, φ(t)z = Y (t; z)}

which is a diffeormorphism that satisfies the group properties φ(0) = I, φ(t + s) = φ(t) ◦ φ(s) for all t, s ∈ R. In particular φ(−t) is the inverse of φ(t).

Therefore, the original domain Ω0 is deformed into the domains

and the boundaries satisfy ∂Ω(t) = φ(t) ∂Ω0. Also, any smooth subdomain W0 of Ω0 is

also deformed into

W(t) = φ(t)W0, t ∈ R

and its boundary is given by ∂W (t) = φ(t)∂W0.

The next results gives geometrical information about the deformations above.

Lemma 2.1 With the above notations, for x0 ∈ ∂Ω0 then φ(t)x0 ∈ Ω(t) and

Dφ(t)(x0)

is an isomorphism in Rn _{that transforms the tangent plane in x}

0 ∈ ∂Ω0, that we denote

Tx0∂Ω0, into the tangent plane to ∂Ω(t) at φ(t)x0, Tφ(t)x0∂Ω(t).

Proof. Just note that if z(s) is a curve in ∂Ω0 with z(0) = x0, then z′(0) = v0 is a

tangent vector at x0 (and conversely). Hence, w(s) = φ(t)(z(s)) is a curve in ∂Ω(t), with

w(0) = y0 and

w′(0) = Dφ(t)(x0)v0

is a tangent vector at ∂Ω(t) . We also recall the following

Definition 2.2 A matrix η(t) is a fundamental matrix of the linear system

X′(t) = A(t)X(t) (2.2)

iff each column of η(t) is a solution of (2.2) and η(t) is nonsingular.

Observe that in particular, η′_{(t) = A(t)η(t). Then we have}

Lemma 2.3 If η(t) is a fundamental matrix of (2.2), then

γ(t) = η−1(t)
∗

= (η∗_(t))−1

is a fundamental matrix of the adjoint system

Y′(t) = −A∗_{(t)Y (t)}

where * denotes the adjoint matrix.

Proof. Differentiate in

η−1(t) ◦ η(t) = I and use (2.2).

Proposition 2.4

i) For x ∈ Rn_{, Dφ(t)x is a fundamental matrix of}

˙

Z(t) = A(t)Z(t)

and Dφ(0) = I, where A(t) = D ~V(φ(t)x).

ii) Denote

|K(x, t)| = det(Dφ(t)x), x_{∈ R}n!‘

then we have the Abel–Liouville–Jacobi formula ∂

∂t|K(x, t)| = tr D ~V
(φ(t)x) |K(x, t)| = div ~V
(φ(t)x) |K(x, t)| hence

|K(x, t)| = eR0tdiv ~V(φ(s)x) ds.

In particular, for t ∈ [−T, T ] there exist C1(T ), C2(T ) such that

0 < C1(T ) ≤ |K(x, t)| ≤ C2(T ) ∀x ∈ Ω0 ∀t ∈ [0, T ]. (2.3)

Remark 2.5 Observe that if W0 ⊂ Ω0 and W (t) = φ(t)W0 then the measure of W (t)

satisfies |W (t)| = Z W(t) 1 dy = Z W0 |K(x, t)| dx = Z W0 eR0tdiv ~V(φ(s)(x)) dsdx. In particular, if div(~V) = 0 then the measure is preserved, that is,

|W (t)| = |W0| ∀ W0 ⊂ Ω0 ∀t ∈ R.

Also, if div(~V) ≤ −d0 <0 at every point, then

|W (t)| ≤ |W0| e−d0t

and we say the flow of (2.1) is contractive.

Finally if div(~V) ≥ d0 >0 at every point, then

|W (t)| ≥ |W0| ed0t

and we say the flow if expansive.

For example for a linear flow, that is, ~V(x) = Mx for a given matrix M, we have

div(~V) = tr(M) =

n

X

i=1

µi= d0

is the trace of M, that is the sum of all eigenvalues of M.

Corollary 2.6 Assume x0 ∈ ∂Ω0 and consider y0 = φ(t)x0 ∈ ∂Ω(t). Then if ~n(x0) is an

unitary outward normal vector to Ω0 at x0 then

N(y0) = ((Dφ(t)x0)∗)−1~n(x0)

is an outward vector at y0. That is, ((Dφ(t))∗x0)−1 is a linear isomorphism in Rn that

transforms the normal space at x0 ∈ ∂Ω0, which we denote, Nx0, into the normal space

to Ω(t) at y0 ∈ ∂Ω(t), which we denote Ny0.

Proof. From Lemma 2.1 a normal vector at y0 = φ(t)(x0) ∈ ∂Ω(t) , ~n, must satisfty

< ~n, Dφ(t)x0~τ >= 0 ∀~τ ∈ Tx0∂Ω0 which reads

<(Dφ(t)x0)∗~n, ~τ >= 0 ∀~τ ∈ Tx0∂Ω0.

Hence we can take ~n such that ((Dφ(t)x0)∗)~n = ~n(x0) which gives the result.

3

Balance equations

The following notations will be used throughout the paper.

Definition 3.1 If for some T > 0, f is defined in

f : ∪t∈(−T,T )Ω(t) × {t} −→ R, (y, t) 7−→ f (y, t)

then we define f in Ω0 as

f : Ω0× (−T, T ) −→ R, f(x, t) = f (φ(t)x, t)

Consider W (t) = φ(t)W0 ⊂ Ω(t), a sufficiently smooth region with boundary ∂W (t).

Then the time variation of the amount of T in W (t) is given by d

dt Z

W(t)

T(y, t) dy

which is computed below. Note that this is the classical Reynolds Transport theorem, [6, 3, 5].

Proposition 3.2 With the notations above, we have that

d dt

Z

W(t)

T(y, t) dy

or _Z W(t) ∂T ∂t(y, t) dy + Z W(t)

divy(T (y, t) . ~V(y)) dy (3.2)

or Z W(t) ∂T ∂t (y, t) dy + Z ∂W(t) T(y, t)~V(y) d~s. (3.3)

Now we will derive the Balance Equations for the quantity T (y, t). In fact we have d dt Z W(t) T(y, t) dy = Z W(t) f(y, t) dy − Z ∂W(t) ~ J d~s

where f (y, t) represents the rate of production/consumption of T per unit volume in

W(t) and ~J is the vector field of the flow of T across the boundary of W (t). Hence the

divercence theorem leads to d dt Z W(t) T(y, t) dy = Z W(t) f(y, t) dy − Z W(t) divyJ dy~ (3.4)

Hence, (3.4) and the Proposition above leads to

Proposition 3.3 Under the assumptions and notations above, the magnitud T satisfies

the balance equations in the moving domains, if and only if the following equivalent con-ditions are satisfied:

∂T

∂t(y, t) + divy(T (y, t) . ~V(y)) = f (y, t) − divy( ~J), y∈ Ω(t), t > 0 (3.5) or

∂

∂tT(x, t) + T (x, t) div(~V)(x, t) = f (x, t) − divy( ~J)(x, t) x∈ Ω0, t > 0. (3.6)

Proof. First, equating (3.2) and (3.4) we get

Z

W(t)

 ∂T

∂t (y, t) + divy(T (y, t)~V(y)) 

dy=

Z

W(t)

(f (y, t) − divyJ~) dy.

Since W (t) = φ(t)(W0), φ(t) is a diffeormorphism and W0 is arbitrary, we get (3.5).

Now, using y = φ(t)x we get in the right hand side of (3.4) Z W0 f(x, t) |K(x, t)| dx − Z W0 divy( ~J)(x, t) |K(x, t)| dx,

4

Boundary and initial conditions

As we consider Dirichlet boundary conditions and using

y= φ(t)x, Ω(t) = φ(t)Ω0, ∂Ω(t) = φ(t)(∂Ω0)

then

T(y, t) = 0 ∀ y ∈ ∂Ω(t) ⇔ T (x, t) = 0 ∀ x ∈ ∂Ω0

As for the initial condition we have, since φ(0) = I,

T(y, 0) = T0(y) ∀ y ∈ Ω0 ⇔ T (x, 0) = T0(x) ∀ x ∈ Ω0.

Thus, (3.5) and (3.6), with boundary and initial conditions read, respectively, 





∂T

∂t(y, t) + divy(T (y, t) . ~V(y)) = f (y, t) − divy( ~J) y∈ Ω(t)

T(y, t) = 0 y ∈ ∂Ω(t) ∀t T(y, 0) = T0(y) y∈ Ω0

(4.1)    ∂ ∂tT(x, t) + T (x, t) div(~V)(x, t) = f (x, t) − divy( ~J)(x, t) x ∈ Ω0 T(x, t) = 0 x ∈ ∂Ω0 ∀t T(x, 0) = T0(x) x∈ Ω0. (4.2) In fact we use (4.2) to define a solution of (4.1), i.e.

T(y, t) satisfies (4.1) ⇔ T (x, t) satisfies (4.2).

5

Balance equations without diffusion

5.1

No flux and no diffusion: pure inertia

With the previous notations, assume divy( ~J) = 0 then the following problems are

equiv-alent

 



∂

∂tT(y, t) + divy(T (y, t) . ~V(y)) = f (y, t) y∈ Ω(t)

T(y, t) = 0 y ∈ ∂Ω(t) ∀t T(y, 0) = T0(y) y ∈ Ω0

Proposition 5.1 With the notations above, (5.1) and (5.2) have a unique explicit solu-tion given by T(y, t) = T0(x) e− Rt 0divyV~(φ(r)x) dr ₊ Z t 0 e−RstdivyV~(φ(r)x) dr f(y, s) ds, y = φ(t)x ∈ Ω(t) and T(x, t) = T0(x) e− Rt 0div ~V(φ(r)x) dr + Z t 0 e−Rstdiv ~V(φ(r)x) dr f(x, s) ds, x∈ Ω 0, respectively.

Proof. The solution of (5.2) is obtained by solving a linear nonhomogeneous ODE

Z′(t) + P (t)Z(t) = h(t), Z(0) = Z0

for each x ∈ Ω0. From this the solution of (5.1) is immediate.

Remark 5.2 Assume in particular that there are no source terms, that is, f = 0. Hence

in (5.1) we have

T(y, t) = T0(x) e− Rt

0divyV~(φ(r)x) dr, y= φ(t)x

Thus, if moreover div(~V) = 0 then

T(y, t) = T0(x) y= φ(t)x,

and T remains constant along the paths of the flow.

On the other hand if the flow is expansive then T (y, t) decreases along the paths of the flow, while it increases if the flow is contractive.

5.2

Flux and no diffusion: transport equations

Below we use ψ(t) = φ−1_{(t) = φ(−t).}

Proposition 5.3 If we assume

~

J(y, t) = ~a(y, t)T (y, t) y∈ Ω(t) with a C1 _{scalar field}

~a_{: R}n_{× R → R}

then the balance equations (4.1) and (4.2) read 





∂

∂tT(y, t) + divy(T (y, t) . ~V(y)) + ∇yT(y, t).~a(y, t) + divy(~a(y, t))T (y, t) = f (y, t) y ∈ Ω(t)

T(y, t) = 0 y∈ ∂Ω(t) ∀t T(y, 0) = T0(y) y ∈ Ω0

and    ∂T ∂t(x, t) + T (x, t)C(x, t) + ∇xT(x, t)~b(x, t) = f (x, t) x∈ Ω0 T(x, t) = 0 x ∈ ∂Ω0 ∀t T(x, 0) = T0(x) x∈ Ω0 (5.4) which are equivalent, where

C(x, t) = divy(~V)(x, t) + divy(~a)(x, t), ~b(x, t) = Dψ(t)y · ~a(y, t).

Proof. Note that (5.3) follows by direct computation from (4.1) using

divy(~a(y, t) T (y, t)) = ∇yT(y, t).~a(y) + T (y, t)divy(~a(y, t)).

On the other hand, for (5.4) we have to write divy(a(y, t)T (y, t)) in terms of x. For

this we observe that since x = ψ(t)y we have T (y, t) = T (ψ(t)y, t) and then ∂T ∂yi (y, t) = n X j=1 ∂T ∂xj (x, t)∂ψj(t)y ∂yi (5.5)

and ∇yT(y, t) = ∇xT(x, t)Dψ(t)y.

Thus, ∇yT(y, t).~a(y, t) = ∇xT(x, t) (Dψ(t)y . ~a(y, t)) and hence

∇yT(y, t).~a(y, t) = ∇xT(x, t)(Dψ(t)y · ~a(y, t)) = ∇xT(x, t)~b(x, t).

Now we show that under some natural geometrical conditions (5.4) (and hence (5.3)) can be solved by using characteristics. Note that we now disregard boundary conditions.

Proposition 5.4 Assume that for all time and y ∈ ∂Ω(t), we have

< ~a(y, t), ~n0(y) >≤ 0

where < ·, · > is the scalar product and and ~n0(y) is the unit outward normal vector at y.

Then (5.3) and (5.4) have a unique solution.

Proof. For (5.4) we use the method of characteristics. Hence, for x0 ∈ Ω0 we define

curves defined on some interval I containing 0

X′(s) = ~b(X(s), t(s)), X(0) = x0

which gives t(s) = s and

X′(t) = ~b(X(t), t), X(0) = x0 ∈ Ω0, (5.6)

which has a solution because ~b ∈ C1_(Rn_).

Hence, from (5.6) and (5.4)

 d

dtZ(t) + C(X(t), t)Z(t) = f (X(t), t)

Z(0) = T0(x0)

whose solution is given by Z(t) = T0(x0)e− Rt 0C(X(r),r) dr + Z t 0 e−RstC(X(r),r) drf(X(s), s) ds. (5.7)

In the computation above we need the solution of (5.6) not to leave Ω0. Thus, if X(t)

reaches the boundary of Ω0 at time t0 at the point y0 = x(t0) ∈ ∂Ω0, the tangent vector

to the characteristic curve at this point is X′

(t0) = ~b(x0, t0), and therefore if it points

inward, that is, if

< ~b(x0, t0), ~n(x0) > ≤ 0 (5.8)

then it will remain in Ω. Note now that from (5.8)

< ~b(x0, t0), ~n(x0) >=< Dψ(t0)y0.~a(y0, t0), ~n(x0) >=< ~a(y0, t0), (Dψ(t0)y0)∗~n(x0) >=

=< ~a(y0, t0), ((Dφ(t0)x0)∗)−1~n(x0) >=< ~a(y0, t0), N(y0) >≤ 0

where we have used Corollary 2.6 and the assumption of this Proposition. With this (5.7) gives the values of the solution in the moving domain.

6

Balance equations with diffusion

Recalling the equivalent equations (4.1) and (4.2) we have

Proposition 6.1 Assume the flux vector field is given by

~

J(y, t) = −k∇yT(y, t) y ∈ Ω(t)

for some k > 0, then (4.1) and (4.2) read 





∂

∂tT(y, t) + ∇yT(y, t).~V(y) + T (y, t)div(~V)(y) − k∆T (y, t) = f (y, t) t∈ Ω(t)

T(y, t) = 0 y∈ ∂Ω(t) ∀t T(y, 0) = T0(y) y ∈ Ω0

and          ∂T(x,t) ∂t + T (x, t) div(~V)(x, t) − k n X k,i=1 ak,i(x, t) ∂2 _{T(x, t)} ∂xk∂xi + n X i=1 ∂T(x, t) ∂xi .si(x, t) ! = f (x, t) T(x, t) = 0 x ∈ ∂Ω0 ∀t T(x, 0) = T0(x) x∈ Ω0 (6.2) where ak,i(x, t) = n X j=1 ∂ψk(t)y ∂yj . ∂ψi(t)y ∂yj = ∇yψk.∇yψi , y= φ(t)x and si(x, t) = n X j=1 ∂2ψi(t)y ∂y2 j = ∆yψi(t)y y= φ(t)x.

Proof. Clearly divy( ~J) = −k∆T (y, t) for y ∈ Ω(t) and we get (6.1). Now for (6.2), we

have from (5.5),

∇yT(y, t) = ∇xT(x, t).Dψ(t)y.

Hence,

divy(−k∇yT(y, t)) = −k divy(∇xT(x, t)Dψ(t)y) =

−k divy n X i=1 ∂ T(x, t) ∂xi . ∂ψi(t)y ∂y1 , ..., n X i=1 ∂ T(x, t) ∂xi . ∂ψi(t)y ∂yn ! . Now observe that

k n X j=1 n X i=1 ∂ T(x, t) ∂xi ∂2ψi(t)y ∂y2 j which leads to −k divy n X i=1 ∂ T(x, t) ∂xi . ∂ψi(t)y ∂y1 , ..., n X i=1 ∂ T(x, t) ∂xi . ∂ψi(t)y ∂yn ! = −k n X k,i=1 ∂2 _{T(x, t)} ∂xk∂xi . n X j=1 ∂ψk(t)y ∂yj . ∂ψi(t)y ∂yj ! − k n X i=1 ∂T(x, t) ∂xi . n X j=1 ∂2_ψ i(t)y ∂y2 j ! . and we get the result.

Concerning the main part in (6.2) we have the following

Proposition 6.2 With the notations above, the term

n X k,i=1 ak,i(x, t) ∂2 T ∂xk∂xi (x, t) can be written in divergence form.

Proof. Just note that

n X k,i=1 ak,i(x, t) ∂2 T(x, t) ∂xk∂xi = n X i=1 ∂ ∂xi n X k=1 ∂T(x, t) ∂xi .ak,i(x, t) ! − n X i=1 ∂T(x, t) ∂xi .ci(x, t) with ci(x, t) = n X k=1 ak,i(x, t) ∂xi

Remark 6.3 Note that now (6.2) can be written as

7

Parabolic PDEs in moving domains

Now we consider general parabolic equations in moving domains. That means that the equations are not necessarily balance equations. Hence, we consider

     ∂ ∂tT(y, t) − k∆yT(y, t) + n X i=1 ∂T ∂yi

(y, t) . gi(y, t) + c(y, t)T (y, t) = f (t, y) y∈ Ω(t)

T(y, t) = 0 y ∈ ∂Ω(t) ∀t T(y, 0) = T0(y) y∈ Ω0

(7.1) with k > 0 and given smooth c(y, t) and ~g(y, t) = (gi(y, t), . . . , gn(y, t)). Note that this

equation contains (6.1) as a particular case.

Then we have the following result whose proof follows from the computation in the sections above.

Proposition 7.1 With the notations above (7.1) is equivalent to

   ∂T ∂t (x, t) − kdiv(B(x, t)) + ∇xT(x, t). ~h(x, t) − ~d(x, t)
+ c(x, t)T (x, t) = f(x, t) x ∈ Ω0 T(x, t) = 0 x ∈ ∂Ω0 ∀t T(x, 0) = T0(x) x∈ Ω0 (7.2) with

B(x, t) = A(x, t)∇xT(x, t), A(x, t) = (ak,i(x, t)), ak,i(x, t) = n X j=1 ∂ψk(t)y ∂yj . ∂ψi(t)y ∂yj , di(x, t) = si(x, t) − ci(x, t), si(x, t) = ∆yψi(t)y, ci(x, t) = n X k=1 ak,i(x, t) ∂xi ~h(x, t) = (~g(x, t) . ∇yψ1(t)y, . . . , ~g(x, t) . ∇yψn(t)y), y= φ(t)x.

Now we are in a position to proof that (7.1) is well posed.

Proposition 7.2 Under the assumptions above, if the initial data satisfies

T0 ∈ L2(Ω0)

then (7.2) and (7.1) have a unique solution.

Proof. Observe that in (7.2)

A(x, t) = Dψ(t)y.(Dψ(t)y)t _{y= φ(t)x.}

Then we show below that this is a positive definite matrix. In fact for ξ ∈ Rn_{, ξ 6= 0, we}

have

< A(x, t)ξ, ξ >=< (Dψ(t)y)t_ξ,(Dψ(t)y)t_{ξ >=}

(Dψ(t)y)tξ

2

since (Dψ(t)y)t _{is non singular. Also, from (2.3), the eigenvalues of Dφ(t) are bounded}

and bounded away from 0 for all t ∈ [0, T ] and so are the eigenvalues of Dψ(t). Therefore there exist α = α(T ) > 0 such that k(Dψ(t)y)t_ξk2 _{≥ α kξk}2_.

Using this, the smoothness of the coefficients and the results in [1, 2], we get that (7.2) has a unique smooth solution and so does (7.1).

8

Maximum principle

In this section we show that the parabolic equations in moving domains possess the maximum principle. We will show this on the particular example of the heat equation

 



∂T

∂t(y, t) − ∆T (y, t) + a(y, t) T (y, t) = 0 y ∈ Ω(t)

T(y, t) = 0 y ∈ ∂Ω(t) ∀t T(y, 0) = T0(y) y∈ Ω0

(8.1) witha a sufficiently smooth coefficient a(y, t). Then we have

Proposition 8.1 With the assumption above, if

T0 ∈ L2(Ω0), T0(x) ≥ 0 x ∈ Ω0

and

α(t) ≤ a(y, t) ∀y ∈ Ω(t) ∀t

for some smooth α(t). Then

T(y, t) ≥ 0, y∈ Ω(t), t≥ 0.

Proof. We multiply (8.1) by the negative part of T , T−_{(y, t), and integrate in Ω(t), to}

get Z Ω(t) ∂T ∂t (y, t) . T −_{(y, t) dy−} Z Ω(t)

∆T (y, t) . T−_{(y, t) dy+}

Z

Ω(t)

a(y, t). T (y, t). T−_{(y, t) dy = 0.}

Using (3.3) for (T−₎2 _{and the fact that T}−_{(y, t) = 0 in ∂Ω(t), because T (y, t) = 0 in}

∂Ω(t), we have 1 2 d dt Z Ω(t) (T−)2(y, t) dy + Z Ω(t)  ∇T−(y, t)   2 dy+ Z Ω(t) a(y, t)(T−)2(y, t) dy = 0. Hence 1 2 d dt T−(., t) 2 L2 (Ω(t))+ α(t) T−(., t) 2 L2 (Ω(t)) ≤ 0 and taking ¯F(t) = kT−_{(., t)k}2 L2_(Ω(t)), we have d dtF¯(t) + 2α(t) ¯F(t) ≤ 0

and Gronwall’s lemma leads to ¯F(t) ≤

T₀− L2 (Ω0)e −2Rt 0α(s) ds = 0, since T− 0 = 0 in Ω0.

9

Energy estimates

In this section we derive suitable energy estimates for the heat equation in a moving domain

 



∂T

∂t(y, t) − ∆T (y, t) + a(y, t) T (y, t) = 0 y ∈ Ω(t)

T(y, t) = 0 y ∈ ∂Ω(t) ∀t T(y, 0) = T0(y) y∈ Ω0

(9.1) with a smooth enough a(y, t). First, we have for nonnegative solutions

Proposition 9.1 Assume

T0 ∈ L2(Ω0) T0(x) ≥ 0, x∈ Ω0

and

α(t) ≤ a(y, t) ∀y ∈ Ω(t) ∀t.

for some smooth α(t) such that lim inf t→∞ 1 t Z t 0 α(s) ds > α0 >0. Then _Z Ω(t) T(y, t) dy ≤ e−R0tα(t) ds Z Ω0 T0(x) dx −−−→ t→∞ 0. Proof. From (3.3) d dt Z Ω(t) T(y, t) dy = Z Ω(t) ∂T ∂t (y, t) dy + Z ∂Ω(t) T(y, t)~V(y) d~s and since T vanishes on the boundary, we have

d dt Z Ω(t) T(y, t) dy = Z Ω(t) ∂T ∂t(y, t) dy. Using this, we integrate in (9.1) in Ω(t), to get

Z Ω(t) ∂T ∂t(y, t) dy − Z Ω(t) ∆T (y, t) dy + Z Ω(t) a(y, t) T (y, t) dy = 0. Now Green’s formula leads to

d dt Z Ω(t) T(y, t) dy − Z ∂Ω(t) ∂T ∂~n(y, t) d~s + Z Ω(t) a(y, t) T (y, t) dy = 0.

By the maximum prinicple we know that T (y, t) ≥ 0 for y ∈ Ω(t) and t ≥ 0, and then

for y ∈ ∂Ω(t) we have ∂T

∂~n(y, t) ≤ 0 and then

Hence, denoting ¯Y(t) = Z Ω(t) T(y, t) dy we have d ¯Y dt(t) + α(t) ¯Y(t) ≤ 0 and Gronwall’s lemma gives

¯ Y(t) = Z Ω(t) T(y, t) dy ≤ e−Rt 0α(t) ds Z Ω0 T0(x) dx −−−→ t→∞ 0. since, by assumption lim inf t→∞ 1 t Z t 0 α(s) ds > α0 >0 and then e−R0tα(s) ds = e −t    1 t Z t 0 α(s) ds    ≤ e−α0t −−−→ t→∞ 0. for t >> 1.

Now without assuming sign on the solutions, we have

Proposition 9.2 With the notations above, assume

T0 ∈ L2(Ω0)

and the function

γ(t) = α(t) − C0(Ω(t)),

is such that for some α1 >0,

lim inf t→∞ 1 t Z t 0 γ(s) ds > α1 >0,

where C0(Ω(t)) is the Poncair`e constant in Ω(t).

Then 0 ≤ Z Ω(t) T2(y, t) dy ≤ e−2R0tγ(s) ds Z Ω(t) T₀2(x) dx −−−→ t→∞ 0.

Proof. Multiply (9.1) by T (y, t) and integrate in Ω(t), to get

Z Ω(t) ∂T ∂t (y, t)T (y, t) dy − Z Ω(t) ∆T (y, t)T (y, t) dy + Z Ω(t) a(y, t) T2(y, t) dy = 0. Using (3.3), the boundary conditions and the Green’s formula we have

Now the Poincar`e inequality in Ω(t) gives for any smooth function vanishing on ∂Ω(t), k∇uk2_L2_(Ω(t)) ≥ C₀(Ω(t)) kuk

2

L2_(Ω(t)). This and the assumption on a(y, t) leads to

1 2 d dtkT (., t)k 2 L2_(Ω(t))+ γ(t) kT (., t)k 2 L2_(Ω(t)) ≤ 0. (9.2) Thus, denoting ¯Z(t) = kT (., t)k2_L2_(Ω(t)), (9.2) reads

d

dtZ(t) + 2γ(t) ¯¯ Z(t) ≤ 0 and Gronwall’s lemma yields

¯ Z(t) ≤ kT0k2L2 (Ω0)e −2Rt 0γ(s) ds −−−→ t→∞ 0. since by assumption lim inf t→∞ 1 t Z t 0 γ(s) ds > α1 >0 and then e−R0tγ(s) ds = e −t    1 t Z t 0 γ(s) ds    ≤ e−α1t −−−→ t→∞ 0 for t >> 1.

References

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[2] Amann, H. Nonhomogeneous linear and quasilinear elliptic and parabolic bound-ary value problems Function spaces, differential operators and nonlinear analysis, (Friedrichroda, 1992), Teubner, 1993, 133, 9-126

[3] Duvaut, G. M´ecanique des milieux continus, Masson, 1990

[4] Hartman, P. Ordinary differential equations, John Wiley and sons, 1964.

[5] Chorin, A. J., Marsden J.E., A mathematical introduction to fluid mechanics, Springer, 1979.